= Solution
{c}
An equilibrium has $y=-x/2$ from the first equation. Substitution in the second gives $-x(x^2+3)/2=0$, so \b[the origin is the only real equilibrium]. Its <Jacobian matrix> is
$$
J=\begin{pmatrix}1&2\\-2&-1\end{pmatrix},\qquad\lambda=\pm i\sqrt3.
$$
Purely imaginary linear <eigenvalues> alone do not classify the nonlinear equilibrium. Use instead
$$
H=x^2+xy+y^2+\frac{x^4}8.
$$
Its derivatives are $H_x=2x+y+x^3/2$ and $H_y=x+2y$, so the vector <field> is $(H_y,-H_x)$. The <chain rule> gives $\dot H=H_xH_y-H_yH_x=0$. Therefore the nonstationary trajectories in the <phase portrait> lie on the level curves
$$
\boxed{x^2+xy+y^2+x^4/8=C.}
$$
The quadratic part is positive definite, with <eigenvalues> $1/2$ and $3/2$, so $H$ has a strict local minimum at zero. Small positive level sets are smooth closed ovals surrounding it, and the vector <field> never vanishes on them. They are periodic orbits. Thus \b[the origin is a nonlinear centre, Lyapunov stable but not asymptotically stable]. The near-origin ellipses have their long axis along $y=-x$. At $(x,0)$ with $x>0$, the velocity points downward, fixing clockwise orientation.
\Image[/past-exam-of-the-algebra-analysis-and-differential-equations-1-course-of-the-university-of-oxford/2009-phase-portrait.png]
{title=Clockwise closed trajectories around the nonlinear centre, given by the conserved level sets}
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